feat: solve No.456

BaffinLee · BaffinLee · commit 00323ba65be0 · 2023-10-02T23:24:14.000+08:00
diff --git a/401-500/456. 132 Pattern.md b/401-500/456. 132 Pattern.md
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+# 456. 132 Pattern
+
+- Difficulty: Medium.
+- Related Topics: Array, Binary Search, Stack, Monotonic Stack, Ordered Set.
+- Similar Questions: .
+
+## Problem
+
+Given an array of `n` integers `nums`, a **132 pattern** is a subsequence of three integers `nums[i]`, `nums[j]` and `nums[k]` such that `i < j < k` and `nums[i] < nums[k] < nums[j]`.
+
+Return `true`** if there is a **132 pattern** in **`nums`**, otherwise, return **`false`**.**
+
+ 
+Example 1:
+
+```
+Input: nums = [1,2,3,4]
+Output: false
+Explanation: There is no 132 pattern in the sequence.
+```
+
+Example 2:
+
+```
+Input: nums = [3,1,4,2]
+Output: true
+Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
+```
+
+Example 3:
+
+```
+Input: nums = [-1,3,2,0]
+Output: true
+Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
+```
+
+ 
+**Constraints:**
+
+
+	
+- `n == nums.length`
+	
+- `1 <= n <= 2 * 105`
+	
+- `-109 <= nums[i] <= 109`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var find132pattern = function(nums) {
+    if (nums.length < 3) return false;
+    var stack = [];
+    var min = Array(nums.length);
+    min[0] = nums[0];
+    for (var i = 1; i < nums.length; i++) {
+        min[i] = Math.min(min[i - 1], nums[i]);
+    }
+    for (var j = nums.length - 1; j >= 0; j--) {
+        if (nums[j] > min[j]) {
+            while (stack.length && stack[stack.length - 1] <= min[j]) stack.pop();
+            if (stack.length && stack[stack.length - 1] < nums[j]) return true;
+            stack.push(nums[j]);
+        }
+    }
+    return false;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).
